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Alina [70]
3 years ago
15

How many grams of zinc phosphate are formed when 48.1 mL of 2.18 M zinc nitrate reacts with excess potassium phosphate?

Chemistry
1 answer:
Roman55 [17]3 years ago
6 0

Answer:

15.4 g of Zn₃(PO₄)₂ are produced

Explanation:

Given data:

Mass of zinc phosphate formed = ?

Volume of zinc nitrate = 48.1 mL (0.05 L)

Molarity of zinc nitrate = 2.18 M

Solution:

Chemical equation:

3Zn(NO₃)₂ + 2K₃PO₄  →   Zn₃(PO₄)₂ + 6KNO₃

Moles of  zinc nitrate:

Molarity = number of moles / volume in litter

Number of moles =  2.18 M × 0.05 L

Number of moles = 0.109 mol

Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:

                     Zn(NO₃)₂           :            Zn₃(PO₄)₂

                         3                    :                 1

                       0.109               :               1/3×0.109 = 0.04 mol

0.04 moles of Zn₃(PO₄)₂ are produced.

Mass of Zn₃(PO₄)₂:

Mass = number of moles × molar mass

Mass = 0.04 mol × 386.1 g/mol

Mass = 15.4 g

   

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General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)

or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

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Answer:

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Kp is defined as the ratio of pressure of products and pressure of reactants:

Kp = \frac{P_{SO_3}^2}{P_{SO_2}^2P_{O_2}}

Kp = \frac{0.0851atm^2}{0.669atm^2*0.395atm}

Kp = 0.0410

I hope it helps!

5 0
3 years ago
Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo
lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, n_i=4

Final orbit of electrons, n_f=2

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

Putting all the values we get :

E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV

We know that : 1\ eV=1.6\times 10^{-19}\ J

So,

E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J

Energy of wave in terms of frequency is given by :

E=hf

f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz

Also, c=f\lambda

\lambda is wavelength

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm

Hence, this is the required solution.

4 0
3 years ago
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